diff --git a/education/math/MATH1210 (calc 1)/Derivatives.md b/education/math/MATH1210 (calc 1)/Derivatives.md
index fcc7342..6b1f876 100644
--- a/education/math/MATH1210 (calc 1)/Derivatives.md	
+++ b/education/math/MATH1210 (calc 1)/Derivatives.md	
@@ -122,6 +122,8 @@ $$ \dfrac{d}{dx}(\dfrac{f(x)}{g(x)}) = \dfrac{f'(x)g(x) -f(x)g'(x)}{(g(x))^2}  $
 $$ \dfrac{d}{dx} e^x = e^x $$
 $$ \dfrac{d}{dx}a^x = a^x*(\ln(a)) $$
 for all $a > 0$
+
+
 # Logarithms
 
 For natural logarithms:
@@ -129,17 +131,27 @@ $$ \dfrac{d}{dx} \ln |x| = \dfrac{1}{x} $$
 
 For other logarithms:
 $$ \dfrac{d}{dx} \log_a x = \dfrac{1}{(\ln a) x}$$
-When solving problems that make use of logarithms, consider making use of logarithmic properties to make life easier, eg:
+When solving problems that make use of logarithms, consider making use of logarithmic properties to make life easier:
 $$ \ln(\dfrac{x}{y}) = \ln(x) - \ln(y) $$
+$$ \ln(a^b) = b\ln(a) $$
+## Logarithmic Differentiation
+This is used when you want to take the derivative of a function raised to a function ($f(x)^{g(x)})$
+
+1. $\dfrac{d}{dx} x^x$
+2. $y = x^x$
+3. $\ln y = \ln x^x$
+4. $\ln(y) = x*\ln(x)$
+5. $\dfrac{d}{dx} \ln y = \dfrac{d}{dx} x \ln x$
+6. $\dfrac{1}{y} \dfrac{dy}{dx} = 1 * \ln x + x * \dfrac{1}{x}$ 
 # Chain Rule
 $$ \dfrac{d}{dx} f(g(x)) = f'(g(x))*g'(x) $$
 ## Examples
 > Given the function $(x^2+3)^4$, find the derivative.
 
 Using the chain rule, the above function might be described as $f(g(x))$, where $f(x) = x^4$, and $g(x) = x^2 + 3)$.
-5. First find the derivative of the outside function function ($f(x) = x^4$):
+7. First find the derivative of the outside function function ($f(x) = x^4$):
 $$ \dfrac{d}{dx} (x^2 +3)^4 = 4(g(x))^3 ...$$
-6. Multiply that by the derivative of the inside function, $g(x)$, or $x^2 + 3$. 
+8. Multiply that by the derivative of the inside function, $g(x)$, or $x^2 + 3$. 
 $$ \dfrac{d}{dx} (x^2 + 3)^4 = 4(x^2 + 3)^3 * (2x)$$ 
 > Apply the chain rule to $x^4$
 
@@ -175,7 +187,7 @@ $$ \dfrac{d}{dx} \cot x = -\csc^2 x $$
 - Given the equation $y = x^2$, $\dfrac{d}{dx} y = \dfrac{dy}{dx} = 2x$.
 
 Given these facts:
-7. Let $y$ be some function of $x$
-8. $\dfrac{d}{dx} x = 1$
-9. $\dfrac{d}{dx} y = \dfrac{dy}{dx}$\
+9. Let $y$ be some function of $x$
+10. $\dfrac{d}{dx} x = 1$
+11. $\dfrac{d}{dx} y = \dfrac{dy}{dx}$\